Q (20)  Prove the following

\small \frac{\sin x - \sin 3x}{\sin^{2}x-\cos^{2}x} = 2\sin x

Answers (1)
S safeer


We know that

 \sin3x = 3\sin x - 4\sin^{3}x \ \ \ , \ \ \cos^{2}-\sin^{2}x = \cos2x \\and \\ \cos2x = 1 - 2\sin^{2}x \\

We use these  identities 

\sin x - \sin3x = \sin x - (3\sin x - 4\sin^{3}x) = 4\sin^{3}x - 2\sin x\\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\sin x (2\sin^{2}x - 1)\\ \\ \cos^{2}x- \sin^{2} = \cos2x\\ \cos2x = 1 - 2\sin^{2}x\sin x - \sin3x = \sin x - (3\sin x - 4\sin^{3}x) = 4\sin^{3}x - 2\sin x\\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\sin x (2\sin^{2}x - 1)\\ \\ \sin^{2}-\cos^{2}x = - \cos2x \ \ \ \ \ \ \ \ \ \ (\cos2x = 1 - 2\sin^{2}x)\\ \sin^{2}-\cos^{2}x = -( 1 - 2\sin^{2}x) = 2\sin^(2)x - 1\\ \\ \frac{\sin x - \sin3x}{\sin^{2}-\cos^{2}x } = \frac{ 2\sin x (2\sin^{2}x - 1)}{ 2\sin^(2)x - 1} = 2\sin x\sin x - \sin3x = \sin x - (3\sin x - 4\sin^{3}x) = 4\sin^{3}x - 2\sin x\\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2\sin x (2\sin^{2}x - 1)\\ \\ \sin^{2}-\cos^{2}x = - \cos2x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \cos2x = 1 - 2\sin^{2}x)\\ \sin^{2}-\cos^{2}x = -( 1 - 2\sin^{2}x) = 2\sin^(2)x - 1\\ \\ \frac{\sin x - \sin3x}{\sin^{2}-\cos^{2}x } = \frac{ 2\sin x (2\sin^{2}x - 1)}{ 2\sin^{2}x - 1} = 2\sin x
                                                                                                       R.H.S.

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